Projective Line - University of Pennsylvania

Projective Line

Where am I?

Celestial Navigation

Earth

Far far away: point at infinity

My location


Earth


My location

Projection of the star


Earth


My location


57 stars for navigation


Earth



My location

Zenith

57 stars for navigation


Earth



My location

rr

r : radius of earth

: angle between stars

Zenith

Celestial Navigation Far far away: point at infinity

My location

Horizon

Sextant

Earth


Zenith


My location

Horizon

Sextant

Earth


Direction does not change because

the star is infinitly far away.

Zenith


My location

Horizon

Sextant

Earth


rr

r : radius of earth


Orthographic proj.

Zenith


Horizon

Sextant

Earth

r : radius of earth



Horizon

Sextant

Earth

At least two stars are needed.


Horizon

Earth


Two points at infinity (vanishing points) tells us about where I am.

Parallel lines in 3D converge to a point in the image.

MLPS-St. Paul International Airport

Indoor point at infinity

3D Parallel Line Projection

Ground plane

Camera plane


Vanishing point

Ground plane

Camera plane


Vanishing point

Ground plane

1. Parallel lines in 3D meet at the same vanishing point in image.

2. The 3D ray passing camera center and the vanishing point is parallel to the lines.

Camera plane

Vanishing Point

Vanishing point

1. Parallel lines in 3D meet at the same vanishing point in image.

2. The 3D ray passing camera center and the vanishing point is parallel to the lines.

3. Multiple vanishing points exist.

Ground plane

Camera plane

Keller Hall

Keller Hall

Vanishing point

Keller Hall

Vanishing point

Vanishing point

Multiple vanishing point

Keller Hall

Vanishing point

Vanishing point

Vanishing line: Horizon

Vanishing line for horizon

Keller Hall

Vanishing point

Vanishing point

What can vanishing line tell us about me?

• Horizon

• Camera pitch angle (looking down)

• Camera roll angle (tilted toward right)

Vanishing line for horizon


Horizon

Earth


Two points at infinity (vanishing points) tells us about where I am.

Parallel 3D planes share the vanishing line.

Different plane produces different vanishing line.

Different plane produces different vanishing line.

How to compute a vanishing point?

Point-Line in Image

A 2D line passing through 2D point (u,v ):

au bv c 0 au bv c 0

( , )uv

Line parameter: (a,b,c)

Point-Line in Image

A 2D line passing through 2D point (u,v ):

au bv c 0 au bv c 0

( , )uv

Line parameter: (a,b,c )

u

a b c v l xT 0

1

where

u

vx

1

a

b

c

land

au bv c 0

Line parameter2D point

=

a

b

c

l2D line:

1

u

vx

Point-Point in ImageA 2D line passing through two 2D points:

1 10 au bv c

1 1( , )u v

1xx2

2 2( , )u v

2 20 au bv c

Point-Point in Image

1 10 au bv c

1 1( , )u v

1xx2

2 2( , )u v

2 20 au bv c

x lT

10 x lT

20

1

1 1

1

x

u

v

2

2 2

1

x

u

v

a

b

c

lwhere

A 2D line passing through two 2D points:


1 10 au bv c

1 1( , )u v

1xx2

2 2( , )u v

2 20 au bv c

x lT

10 x lT

20

1

1 1

1

x

u

v

2

2 2

1

x

u

v

a

b

c

lwhere


xl 0

x

T

1

T

2

2x3

00

A


1 10 au bv c

1 1( , )u v

1xx2

2 2( , )u v

2 20 au bv c

x lT

10 x lT

20

1

1 1

1

x

u

v

2

2 2

1

x

u

v

a

b

c

lwhere


xl 0

x

T

1

T

2

3x2

00

A

null A l x x1 2

=or


(1804,934)(1052,1323)

function GetLineFromTwoPoints

x1 = [1804;934;1];x2 = [1052;1323;1];

l

1 1( , )u v

1xx2

2 2( , )u v


(1804,934)(1052,1323)


x1 = [1804;934;1];x2 = [1052;1323;1];

l = Vec2Skew(x1)*x2;

Cross product

a b a b

a b ab

ab a b

a a b

a -a b

-a a b

a b

a b

2 3 3 2

3 1 1 3

1 2 2 1

3 2 1

3 1 2

2 1 3

-

= -

-

0 -

0

0

Cross product with skew-symmetric matrix representation:

function skew = Vec2Skew(v)skew = [0 -v(3) v(2);

v(3) 0 -v(1);-v(2) v(1) 0];

l

1 1( , )u v

1xx2

2 2( , )u v


(1804,934)(1052,1323)


x1 = [1804;934;1];x2 = [1052;1323;1];

l = Vec2Skew(x1)*x2;

l =

-389-752

1404124

Cross product

a b a b

a b ab

ab a b

a a b

a -a b

-a a b

a b

a b

2 3 3 2

3 1 1 3

1 2 2 1

3 2 1

3 1 2

2 1 3

-

= -

-

0 -

0

0

Cross product with skew-symmetric matrix representation:

function skew = Vec2Skew(v)skew = [0 -v(3) v(2);

v(3) 0 -v(1);-v(2) v(1) 0];

l

1 1( , )u v

1xx2

2 2( , )u v

Two 2D lines in an image intersect at a 2D point:

Line-Line in Image

1

1 1

1

l

a

b

c

( , )uvx

2

2 2

2

l

a

b

c

1 1 10 au bv c

2 2 20 a u bv c


Line-Line in Image

l xT

10 l xT

20

1

1 1

1

l

a

b

c

( , )uvx

2

2 2

2

l

a

b

c

1 1 10 au bv c

2 2 20 a u bv c

1

x

u

v

1

1 1

1

l

a

b

c

where

2

2 2

2

l

a

b

c


Line-Line in Image

l xT

10 l xT

20

lx 0

l

T

1

T

2

l x x1 2

=

1

1 1

1

l

a

b

c

( , )uvx

2

2 2

2

l

a

b

c

1 1 10 au bv c

2 2 20 a u bv c

1

x

u

v

1

1 1

1

l

a

b

c

where

2

2 2

2

l

a

b

c

2x3

00

A

null A

or

Line-Line in Image

1

1 1

1

l

a

b

c

( , )uvx

2

2 2

2

l

a

b

c

function GetPointFromTwoLines

l1 = [-398;-752;1404124];l2 = [310;-924;303790];x = Vec2Skew(l1)*l2;x = x/x(3)

x =

1779.0925.61

similar to (1804,934)

2D Point and Line Duality

l x x1 2

The intersection between two lines:

x l l1 2

Given any formula, we can switch the

meaning of point and line to get

another formula.

The 2D line joining two points:


l x x1 2


x l l1 2

Given any formula, we can switch the meaning of

point and line to get another formula.

1x

2x


T: Transformation-T

2 1 2 1 x Tx l T l


l x x1 2


x l l1 2



1x

2x


T

T T -1 -T T

1 1 1 1 1 1 2 2 l x l T Tx T l Tx l x

T: Transformation-T

2 1 2 1 x Tx l T l

0


l x x1 2


x l l1 2

Given any formula, we can switch the meaning

of point and line to get another formula.

1x

2x


T

T T -1 -T T

1 1 1 1 1 1 2 2 l x l T Tx T l Tx l x

T: Transformation-T

2 1 2 1 x Tx l T l


l x x1 2


x l l1 2



2 1

cos sin 0

-sin cos 0

0 0 1

x x

1x

2x


T

T T -1 -T T

1 1 1 1 1 1 2 2 l x l T Tx T l Tx l x

T: Transformation

T

2 1 1

cos sin 0 cos sin 0

-sin cos 0 -sin cos 0

0 0 1 0 0 1

l l l

-T

2 1 2 1 x Tx l T l

?


l x x1 2


x l l1 2

Given any formula, we can switch the meaning

of point and line to get another formula.

2 1

cos sin 0

-sin cos 0

0 0 1

x x

1x

2x


T

T T -1 -T T

1 1 1 1 1 1 2 2 l x l T Tx T l Tx l x

T: Transformation

T

2 1 1

cos sin 0 cos sin 0

-sin cos 0 -sin cos 0

0 0 1 0 0 1

l l l

-T

2 1 2 1 x Tx l T l

Geometric Interpretation (Point)

CO x

x


CO x

x

-1K x


CO

x̂

-1ˆ x K xx̂

Normalized coordinate:

-1ˆ x K x


CO

1x̂

-1

1 1ˆ x K x

1x̂

2x̂

-1

2 2ˆ x K x

2x̂

-1

2 2ˆ x K x


-1

1 1ˆ x K x

Geometric Interpretation (Line)

CO

1x̂

-1

1 1ˆ x K x

1x̂

2x̂

-1

2 2ˆ x K x

2x̂

l̂

l̂

-1

2 2ˆ x K x


-1

1 1ˆ x K x

-T

-1 Tˆ l K l K l due to duality

1ˆ ˆl = x 2

ˆx

where ?

Geometric Interpretation (Line)

CO

1x̂

-1

1 1ˆ x K x

1x̂

2x̂

-1

2 2ˆ x K x

2x̂

l̂

l̂

-1

2 2ˆ x K x


-1

1 1ˆ x K x

-T


1ˆ ˆl = x 2

ˆx

where

CO

1x̂

2x̂

l̂

-1

1 1ˆ x K x

1x̂

2x̂

-1

2 2ˆ x K x

l̂

Geometric Interpretation (Line)3D planeπ

-1

2 2ˆ x K x


-1

1 1ˆ x K x

-T


1ˆ ˆl = x 2

ˆx

where

A 2D line in an image defines to a 3D plane

passing the camera center:

l̂ π

CO

1x̂

2x̂

l̂

-1

1 1ˆ x K x

1x̂

2x̂

-1

2 2ˆ x K x

l̂


-1

2 2ˆ x K x


-1

1 1ˆ x K x

-T


1ˆ ˆl = x 2

ˆx

where



l̂ π

Plane normal: 1 1ˆ x 2 2

ˆˆ x l

l̂

?

CO

1x̂

2x̂

l̂

-1

1 1ˆ x K x

1x̂

2x̂

-1

2 2ˆ x K x

l̂


-1

2 2ˆ x K x


-1

1 1ˆ x K x

-T


1ˆ ˆl = x 2

ˆx

where



l̂ π


ˆˆ x l

l̂

CO

1x̂

-1

2 2ˆ x K x


-1

1 1ˆ x K x

2x̂

l̂

-T


1ˆ ˆl = x 2

ˆx

where

-1

1 1ˆ x K x

1x̂

2x̂

-1

2 2ˆ x K x

l̂




l̂ π


ˆˆ x l

ˆ

0

lπ

ˆ

0

l

l̂

CO

l̂

Geometric Interpretation (Line-Line)

1π

CO


l1ˆl2

ˆ

1π

2π

CO


l1ˆl2

ˆ

1π

2π

x̂

x̂

2D lines in an image intersect a 2D point corresponding to a 3D ray:

x̂ 1ˆ= l 2

l̂

CO


l1ˆl2

ˆ

1π

2π

x̂

x̂

2D lines in an image intersect a 2D point corresponding to a 3D ray:

x̂ 1ˆ= l 2

l̂1l̂

2l̂ ˆx

: the 3D ray is perpendicular to two plane normals.

Projective Line - University of Pennsylvania

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